# Why does it matter whether or not an ammeter has zero resistance?

1. Jan 9, 2013

### question dude

Apparently an ideal ammeter has zero resistance.

But for a series circuit, why does it matter?

The purpose of an ammeter is to give the reading for the amount of current going round a 'loop' of circuit. Having some resistance does not prevent the ammeter from giving a true reading, right?

So if this ammeter is connected with a resistor, and the ammeter gives a reading of 10A, that is the true value of current going through the resistor. So if you're trying to do an experiment to get the IV (current-voltage) curve of a component, the plotted curve would look the exactly the same regardless of whether you've used a zero resistance ammeter or not, right?

2. Jan 9, 2013

### VantagePoint72

Because V=IR. Circuits most commonly run at a fixed voltage (i.e. "a 6-volt battery"). The current flowing is then determined by the resistance of the circuit. Since the ammeter has its own very small resistance, this will change its reading somewhat when put in series with your resistors. I.e., instead of reading I = V/Rresistors, it will read I = V/(Rresistors + Rammeter). In the idealization of Rammeter=0, these are equal.

3. Jan 9, 2013

### question dude

I still don't understand why it matters whether the ammeter has zero resistance if you're trying to measure the current and voltage passing through a component (construct an IV curve for this component)

Sure, I understand that current is reduced from what it 'should' be (it would be less than if there was only one resistor/component in the circuit). However, the actual current flowing through the circuit will still be reflected by the reading on the ammeter - is this right or wrong? and if thats right, then surely the IV curve is not affected

4. Jan 9, 2013

### VantagePoint72

I'm not sure which part of my first response wasn't clear, so you're going to have to be more specific with your question. If all your resistors are in a series, then an ammeter gives the current of the entire circuit, not any one piece. Are you perhaps confusing an ammeter's mode of operation with a voltmeter, where the latter can tell you the voltage drop across a single resistor?

Again, I'm afraid I don't understand what it is you're asking. Your two sentences contradict each other. Introducing the ammeter into the circuit causes the current drop slightly. Hence, the reading it gives is the not exactly the same as the current flowing through the circuit without the ammeter. You say, "However, the actual current flowing through the circuit will still be reflected by the reading on the ammeter"—yes, the actual current flowing through the circuit when it includes the ammeter. That's not the value we're after, since the ammeter isn't a permanent component of the circuit.

5. Jan 9, 2013

### grzz

Plotting an I-V curve for a component involves measuring both the current and the voltage across the component.

If the voltage of the component is measured by a voltmeter placed across the component, then ammeter will not give the current through the component but the current through both the voltmeter and the component.

If the ammeter is to measure only the current through the component, then the voltmeter will have to be placed across both the component and the ammeter. Hence the voltage recorded will not be that across the component - unless the ammeter has a much smaller resistance when compared with that of the component.

6. Jan 9, 2013

### Staff: Mentor

Sometimes it helps to imagine exaggerated circumstances in order to "see" an effect.

Suppose your bargain-basement ammeter has an internal resistance of 100Ω. You have a circuit consisting of a 10V battery and a 1Ω resistor for which you have calculated that the current in the circuit should be 10V/1Ω = 10 A. You decide to insert your ammeter to find out if this is truly the case. What will the ammeter read? Does this ammeter let you see how the circuit normally operates?

7. Jan 9, 2013

### question dude

10v/(1 + 100), which means current would be about 100 times smaller than without the ammeter in place, I get that.

but in a circuit investigating the IV (current and voltage) characteristuc of a component, there's a variable resistor that will be used to vary to voltage (supplied to the tested component) or current (for the whole circuit. Then the figures for voltage and current will be plotted in an IV graph.

take a look at the whole circuit:

Variable resistor is shown by the rectangle with an arrow, the component tested is the green rectangle.

so why does it matter that the ammeter has a little bit of fixed resistance? because the resistance of the circuit will be changed anyway in steps (by the variable resistor) to measure voltage and current. Do you all get what I mean, I'm trying to say that it doesn't matter that the current reading on the ammeter doesn't give a figure for when only the tested component is in the circuit. What really matters for this circuit is that the ammeter shows the correct reading of the amount of current that flows through the component.

Last edited: Jan 9, 2013
8. Jan 9, 2013

### Staff: Mentor

You've presented a particular instance in which the impedance of the ammeter is not a factor in the proper (desirable) operation of the circuit.

But your opening statement of the "problem" left considerably more leeway for interpretation; presumably an "ideal ammeter" is one that would be useful in any circuit, not just particular circuits where its changing the original circuit operation by its insertion doesn't matter.

9. Jan 9, 2013

### VantagePoint72

It will be changed in known steps. The resistance of the ammeter is unknown and means all your current measurements in the IV plot will be offset by their true value for every voltage reading. You began by asking: "So if this ammeter is connected with a resistor, and the ammeter gives a reading of 10A, that is the true value of current going through the resistor," and the answer to that is no, whatever other stuff you want to throw into your circuit.

You've received the same answer from several people now. If you're convinced that we're not answering the question you mean us to be answering, then I can only say you've done a very poor job of explaining your question.

10. Jan 9, 2013

### grzz

The ammeter in the circuit measures the current through the required component plus the current through the voltmeter.

11. Jan 9, 2013

### CWatters

What gneill said. In some cases it doesn't matter but in most cases it does.

Regarding your circuit... Suppose that instead of characterising the device in green you were using the meters and potentiometer to set the current and voltage to a particular value. If the ameter has internal resistance when you removed it from the circuit the current would change. To set the current to the right value you would need to know what the meter resistance was, calculate how that affects the current and set it accordingly. That's not too difficult when it's just a resistor but what happens when the device is an active component? In some cases inserting the ameter might turn the device off so the current can't be set.

Your circuit shows both a volt meter and a current meter. Have you considered that the input impedance of the voltmeter also affects the current meter reading? If the voltmeter is also non-idea your charaterisation of the device in green will also be off.

We're pretty lucky. My father was an engineer in the days when meters didn't have FET input stages or fancy electronics to allow very low resistance shunts to be used. In those days the meter impedance wasn't even constant, it changed depending on the range it was set to.